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Chapter 15: Problem 1

What are the formal oxidation states of \(\mathrm{N}\) or \(\mathrm{P}\) in the following species? (a) \(\mathrm{N}_{2} ;(\mathrm{b})\left[\mathrm{NO}_{3}\right]^{-} ;(\mathrm{c})\left[\mathrm{NO}_{2}\right]^{-}\) (d) \(\mathrm{NO}_{2} ;\) (e) \(\mathrm{NO} ;(\mathrm{f}) \mathrm{NH}_{3} ;(\mathrm{g}) \mathrm{NH}_{2} \mathrm{OH} ;(\mathrm{h}) \mathrm{P}_{4} ;(\mathrm{i})\left[\mathrm{PO}_{4}\right]^{3-}\) (j) \(P_{4} O_{6} ;(k) P_{4} O_{10}\)

Short Answer

Expert verified
N: (a) 0, (b) +5, (c) +3, (d) +4, (e) +2, (f) -3, (g) -1; P: (h) 0, (i) +5, (j) +3, (k) +5.

Step by step solution

01

Understanding Oxidation States

The oxidation state is a measure of the degree of oxidation of an atom in a substance. It is defined as the charge an atom would have if electrons were completely transferred from an atom with lower electronegativity to an atom with higher electronegativity.
02

Rules for Calculating Oxidation States

1. The oxidation state of an element in its pure form is 0. 2. The sum of the oxidation states of all atoms in a neutral compound is 0. 3. The sum of the oxidation states of all atoms in an ion is equal to the charge of the ion. 4. Usually, Group 1 elements have +1, Group 2 have +2, hydrogen is +1, and oxygen is -2, except in some peroxides and superoxides.
03

Calculating Oxidation State of N in N2

For \(\text{N}_2\), the oxidation state of each nitrogen atom is 0, as nitrogen is in its elemental form.
04

Calculating Oxidation State of N in NO3^-

In \([\text{NO}_3]^−\), oxygen has an oxidation state of -2. Let the oxidation state of nitrogen be x:\[x + 3(-2) = -1\], solving yields \x = +5\.
05

Calculating Oxidation State of N in NO2^-

In \([\text{NO}_2]^−\), oxygen is -2. Let the oxidation state of nitrogen be x:\[x + 2(-2) = -1\], solving yields \x = +3\.
06

Calculating Oxidation State of N in NO2

In \(\text{NO}_2\), oxygen is -2. Let the oxidation state of nitrogen be x:\[x + 2(-2) = 0\], solving yields \x = +4\.
07

Calculating Oxidation State of N in NO

In \(\text{NO}\), oxygen is -2. Let the oxidation state of nitrogen be x:\[x - 2 = 0\], solving yields \x = +2\.
08

Calculating Oxidation State of N in NH3

In \(\text{NH}_3\), hydrogen is +1. Let the oxidation state of nitrogen be x:\[x + 3(+1) = 0\], solving yields \x = -3\.
09

Calculating Oxidation State of N in NH2OH

In \(\text{NH}_2\text{OH}\), hydrogen is +1 and oxygen is -2. Let x be the oxidation state of nitrogen:\[x + 2(+1) - 2 + 1 = 0\], solving yields \x = -1\.
10

Calculating Oxidation State of P in P4

For \(\text{P}_4\), the oxidation state of each phosphorus atom is 0, as phosphorus is in its elemental form.
11

Calculating Oxidation State of P in PO4^3-

In \([\text{PO}_4]^{3-}\), oxygen is -2. Let the oxidation state of phosphorus be x:\[x + 4(-2) = -3\], solving yields \x = +5\.
12

Calculating Oxidation State of P in P4O6

In \(\text{P}_4\text{O}_6\), oxygen is -2. Let x be the oxidation state of phosphorus:\[4x + 6(-2) = 0\], solving yields \x = +3\.
13

Calculating Oxidation State of P in P4O10

In \(\text{P}_4\text{O}_{10}\), oxygen is -2. Let x be the oxidation state of phosphorus:\[4x + 10(-2) = 0\], solving yields \x = +5\.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inorganic Chemistry
Inorganic chemistry is the branch of chemistry that deals with the properties and behavior of inorganic compounds. This broad category encompasses all chemical compounds except organic compounds, which are carbon-based.

Inorganic compounds often consist of a metallic element, a non-metallic element, the combination of both, or other chemical species. They are commonly found in many areas, including the earth's crust, industrial processes, and in biological systems.

One of the key concepts in inorganic chemistry is understanding oxidation states. Oxidation states, also known as oxidation numbers, refer to the hypothetical charges that atoms would have if all bonds to different atoms were 100% ionic. Mastering this idea allows chemists to determine how electrons are exchanged between different elements in chemical reactions. This understanding is crucial in analyzing complex inorganic reactions.
Nitrogen Oxidation States
The oxidation states of nitrogen can vary significantly. Nitrogen is capable of forming a variety of compounds by exhibiting oxidation states ranging from -3 to +5. This property makes it extremely versatile and reactive in chemical processes.- In molecular nitrogen \( ext{N}_2\), each nitrogen atom has an oxidation state of 0 since it is in its elemental form.- In the ion \([ ext{NO}_3]^-\), nitrogen achieves its highest oxidation state of +5. Here, nitrogen is bonded to three oxygen atoms each with an oxidation state of -2, and the overall charge of the ion is -1.- In \( ext{NH}_3 \), nitrogen takes an oxidation state of -3, its lowest state, because it forms three covalent bonds with hydrogen atoms (+1 each).By determining these oxidation states, students can predict how nitrogen behaves in various chemical environments. Understanding the oxidation state helps elucidate electron transfer among atoms involved in reactions, essential for grasping advanced inorganic chemistry topics.
Phosphorus Oxidation States
Phosphorus, similar to nitrogen, can demonstrate multiple oxidation states, though it does not extend as widely. Typically, phosphorus is found in oxidation states ranging from -3 to +5.- In \( ext{P}_4\), elemental phosphorus, the oxidation state is 0 because it is in its natural state in a tetrahedral form.- In \([ ext{PO}_4]^{3-}\), phosphorus is in its highest oxidation state of +5. Here, phosphorus is surrounded by four oxygen atoms, each with an oxidation state of -2.- Compounds like \( ext{P}_4 ext{O}_6\) exhibit phosphorus in the +3 oxidation state, which is relatively common and indicates the phosphorus atoms are covalently bonded to oxygen atoms.Phosphorus compounds are essential in biological systems as well. For example, phosphates play crucial roles in energy transfer and genetic material structure.
Chemical Species Analysis
Chemical species analysis involves understanding, identifying, and predicting the chemical properties and reactions of different species. It builds on identifying the oxidation states of the chemical elements involved.

A methodical approach is essential when analyzing compounds and ions:- Break down the compound to determine the individual oxidation states, considering the rules which guide these calculations, such as knowing the oxidation states of oxygen typically are \(-2\), and hydrogen is usually \(+1\).- Sum the oxidation states within a molecule or ion to help balance and identify the correct states involved.Through this analytical process, students can not only solve textbook problems but also apply these skills to real-world scenarios. By systematically determining oxidation states, chemists predict reactions and aid in developing new chemical compounds and materials for various applications.

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Most popular questions from this chapter

Electron diffraction and spectroscopic studies of mixed fluoro/chloro phosphorus pentahalides are consistent with trigonal bipyramidal structures in which the most electronegative halogens occupy the axial positions. Confirm that this statement is in agreement with \(\mathrm{PCl}_{3} \mathrm{F}_{2}\) \(\mathrm{PCl}_{2} \mathrm{F}_{3}\) and \(\mathrm{PClF}_{4}\) having \(D_{3 \mathrm{h}}, C_{2 \mathrm{v}}\) and \(C_{2 \mathrm{v}}\) symmetries, respectively. Draw the structure of each compound and state whether the compound is polar.

(a) Discuss structural variation among the phosphorus(III) and phosphorus(V) halides, indicating where stereochemical non-rigidity is possible. (b) On what basis is it appropriate to compare the lattice of \(\left[\mathrm{PCl}_{4}\right]\left[\mathrm{PCl}_{6}\right]\) with that of \(\mathrm{CsCl}\) ?

Comment on the fact that \(\mathrm{AlPO}_{4}\) exists in several forms, each of which has a structure which is also that of a form of silica.

(a) Predict the \(^{31} \mathrm{P} \mathrm{NMR}\) spectrum of \(\left[\mathrm{HPF}_{5}\right]^{-}\) (assuming a static structure) given that \(J_{\mathrm{PH}}=939 \mathrm{Hz}\) \\[ J_{\mathrm{PF}(\text { axial })}=731 \mathrm{Hz} \text { and } J_{\mathrm{PF}(\text { cquatorial })}=817 \mathrm{Hz} \\] (b) The \(\left[\mathrm{BiF}_{7}\right]^{2-}\) and \(\left[\mathrm{SbF}_{6}\right]^{3-}\) ions have pentagonal bipyramidal and octahedral structures, respectively. Are these observations consistent with VSEPR theory? (c) Consider the following reaction scheme (K.O. Christe \((1995) J . A m .\) Chem. Soc., vol. \(117,\) p. 6136 ): \\[ \mathrm{NF}_{3}+\mathrm{NO}+2 \mathrm{SbF}_{5} \quad \stackrel{420 \mathrm{K}}{\longrightarrow} \quad\left[\mathrm{F}_{2} \mathrm{NO}\right]^{+}\left[\mathrm{Sb}_{2} \mathrm{F}_{11}\right]^{-}+\mathrm{N}_{2} \\] Discuss the reaction scheme in terms of redox and Lewis acid-base chemistry. Comment on the structures of, and bonding in, the nitrogen-containing species in the scheme.

(a) We noted that \(\left[\mathrm{N}_{3}\right]^{-}\) is isoelectronic with \(\mathrm{CO}_{2}\), Give three other species that are also isoelectronic with \(\left[\mathrm{N}_{3}\right]^{-}\) (b) Describe the bonding in \(\left[\mathrm{N}_{3}\right]^{-}\) in terms of an \(\mathrm{MO}\) picture.
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